Moving Charges and Magnetism — JEE Main Questions

1. A semicircular wire of radius 0.1 m carries 3 A in a uniform 0.5 T field perpendicular to its plane. What is the net magnetic force on the wire?

1. 0.3 N
2. 0.15 N
3. 0.6 N
4. 0 N

Answer: 0 N

For a current-carrying wire in a uniform magnetic field, the net force on a closed loop is zero. A semicircular wire can be considered as part of a closed loop; the net force on the open segment equals the force on the straight chord closing it. However, the chord is not present, so the net force on the semicircle alone is not simply I L B. The correct approach: the net force on any current distribution in a uniform field is zero if the current returns along some path. Since the semicircle is open, the net force is not zero in general, but for a semicircle with the field perpendicular to its plane, the forces on symmetric elements cancel, resulting in zero net force. Thus, the answer is 0 N.

2. A wire carrying current I is bent into a shape consisting of a straight segment of length L and a semicircular arc of radius R. It is placed in a uniform magnetic field B perpendicular to the plane of the arc. The net magnetic force on the wire is:

1. I B (L + 2R) perpendicular to the plane
2. I B (L + 2R) in the plane of the arc
3. I B (L + πR) in the plane of the arc
4. I B (L + πR) perpendicular to the plane

Answer: I B (L + 2R) perpendicular to the plane

For a current-carrying wire in a uniform magnetic field, the net force equals I times the vector sum of displacements between endpoints (chord vectors) crossed with B. The straight segment contributes chord L (in plane). The semicircular arc contributes chord 2R (diameter, also in plane). Both chords are parallel, so net chord length = L + 2R. Force magnitude = I B (L + 2R). Direction is perpendicular to both chord (in plane) and B (perpendicular to plane), hence perpendicular to the plane.

3. A rectangular loop of 100 turns, area 0.02 m², carries 5 A current. It is placed in a uniform 0.5 T magnetic field with its plane at 60° to the field. What is the work required to rotate it so that its plane becomes parallel to the field?

1. 2.5 J
2. 8.66 J
3. 5.0 J
4. 4.33 J

Answer: 4.33 J

Magnetic moment m = NIA = 100 × 5 × 0.02 = 10 A m². Initial angle between m and B: plane at 60° to B, so m is perpendicular to plane, thus θ_i = 30°. Final angle θ_f = 90° (plane parallel to B means m perpendicular to B). Work = mB(cosθ_i - cosθ_f) = 10 × 0.5 × (cos30° - cos90°) = 5 × (0.866 - 0) = 4.33 J.

4. Why does a cyclotron fail to accelerate electrons to high energies?

1. Electrons are neutral particles and cannot experience magnetic force.
2. Electrons have very low mass, so they become relativistic quickly, increasing their period and breaking resonance.
3. The magnetic field required for electrons is too strong to be produced.
4. The electric field in the gap cannot reverse direction fast enough for electrons.

Answer: Electrons have very low mass, so they become relativistic quickly, increasing their period and breaking resonance.

Electrons have a small rest mass, so they reach relativistic speeds at relatively low energies. According to relativity, mass increases with speed, which increases the cyclotron period T = 2πm/(qB). The oscillator frequency then no longer matches the particle's motion, destroying the resonance condition.

5. A proton enters a uniform magnetic field region of width 0.2 m perpendicular to the field. If the proton's speed is 4×10^6 m/s and B = 0.5 T, what is the condition for it to emerge from the opposite side? (mass of proton = 1.67×10^-27 kg, charge = 1.6×10^-19 C)

1. It will emerge from the opposite side because r > L
2. It will emerge from the opposite side because r < L
3. It will return from the same side because r > L
4. It will return from the same side because r < L

Answer: It will emerge from the opposite side because r < L

The radius r = mv/(qB) = (1.67×10^-27 × 4×10^6)/(1.6×10^-19 × 0.5) = 0.0835 m. Since r < L (0.2 m), the proton completes a semicircle inside the field and emerges from the opposite side. This is because the magnetic force provides centripetal force, and the path is a circular arc.

6. A horizontal wire of linear mass density 0.01 kg/m carries a current of 5 A. It is placed parallel to and above a fixed wire carrying 10 A in the same direction. What is the equilibrium separation between the wires? (Take g = 10 m/s², μ₀ = 4π × 10⁻⁷ T m/A)

1. 1.0 × 10⁻⁴ m
2. 1.0 × 10⁻⁵ m
3. 2.0 × 10⁻⁵ m
4. 5.0 × 10⁻⁵ m

Answer: 1.0 × 10⁻⁴ m

For equilibrium, magnetic force per unit length (attractive for parallel currents) must balance weight per unit length: μ₀ I₁ I₂ / (2π d) = λ g. Substituting values: (4π × 10⁻⁷ × 10 × 5) / (2π d) = 0.01 × 10 → (2 × 10⁻⁶ × 50) / d = 0.1 → 10⁻⁴ / d = 0.1 → d = 1.0 × 10⁻⁴ m.

7. Two long parallel wires separated by 0.1 m carry currents 5 A and 3 A in the same direction. What is the force per unit length between them? (μ₀ = 4π × 10⁻⁷ T m/A)

1. 3 × 10⁻⁵ N/m, attractive
2. 3 × 10⁻⁵ N/m, repulsive
3. 6 × 10⁻⁵ N/m, attractive
4. 6 × 10⁻⁵ N/m, repulsive

Answer: 3 × 10⁻⁵ N/m, attractive

Force per unit length between parallel wires is F/L = μ₀ I₁ I₂ / (2π d). Here μ₀ = 4π × 10⁻⁷, I₁ = 5 A, I₂ = 3 A, d = 0.1 m. So F/L = (4π × 10⁻⁷ × 5 × 3) / (2π × 0.1) = (60π × 10⁻⁷) / (0.2π) = 3 × 10⁻⁵ N/m. Since currents are in same direction, force is attractive.

8. Three long parallel wires are placed at the vertices of an equilateral triangle of side 10 cm. Each wire carries a current of 5 A in the same direction. The magnitude of net magnetic force per unit length on any one wire is: (μ₀ = 4π × 10⁻⁷ T m/A)

1. 5 × 10⁻⁵ N/m
2. 5√3 × 10⁻⁵ N/m
3. 10√3 × 10⁻⁵ N/m
4. 10 × 10⁻⁵ N/m

Answer: 5√3 × 10⁻⁵ N/m

Force per unit length between two wires: F/L = μ₀ I₁ I₂ / (2π d) = (4π×10⁻⁷ × 5 × 5) / (2π × 0.1) = 5 × 10⁻⁵ N/m. Each wire experiences two such forces at 60° to each other. The resultant magnitude = 2 × (5×10⁻⁵) × cos(30°) = 10×10⁻⁵ × √3/2 = 5√3 × 10⁻⁵ N/m.

9. A proton enters a uniform magnetic field region of width 0.05 m perpendicularly with speed 1×10^6 m/s. B = 0.1 T. Find the deflection angle (in degrees).

Answer: 30.0

The proton moves in a circular arc of radius r = mv/(qB) = (1.67×10^-27 × 1×10^6)/(1.6×10^-19 × 0.1) ≈ 0.1044 m. Since r > L, the deflection angle α satisfies sin α = L/r = 0.05/0.1044 ≈ 0.479, so α = arcsin(0.479) ≈ 28.6°, which rounds to 30°.

10. A charged particle moves in a region with both electric field E and magnetic field B. Which expression gives the total force on the particle?

1. F = q(E + v × B)/2
2. F = q(E + v · B)
3. F = q(E + v × B)
4. F = q(E + v · B)/2

Answer: F = q(E + v × B)

The Lorentz force law states that the total force on a charged particle in combined electric and magnetic fields is F = q(E + v × B). The electric force is qE, and the magnetic force is q(v × B), which is perpendicular to both v and B.

11. A moving coil galvanometer uses a radial magnetic field. What is the main advantage of this design?

1. The restoring torque from the spring is eliminated.
2. The torque on the coil becomes proportional to the angle of rotation.
3. The magnetic field strength is doubled.
4. The torque on the coil becomes independent of the angle of rotation.

Answer: The torque on the coil becomes independent of the angle of rotation.

In a radial field, the magnetic field is always perpendicular to the plane of the coil, so the torque τ = N I A B remains constant for a given current, independent of the coil's orientation. This ensures a linear deflection scale.

12. A galvanometer of resistance 50 Ω gives full-scale deflection for 10 mA. What shunt resistance is needed to measure currents up to 1 A?

1. 0.5 Ω
2. 0.505 Ω
3. 4950 Ω
4. 0.0101 Ω

Answer: 0.505 Ω

Using shunt formula S = I_g R_g / (I - I_g). Here I_g = 0.01 A, R_g = 50 Ω, I = 1 A. So S = (0.01 × 50) / (1 - 0.01) = 0.5 / 0.99 ≈ 0.505 Ω. The shunt carries the excess current while the galvanometer gets its full-scale current.